A teacher gives a group 12 identical cells and a heating coil…

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A teacher gives a group $12$ identical cells and a heating coil. Each cell has emf $1.5\ \text{V}$ and internal resistance $0.50\ \Omega$ . The coil has resistance $2.0\ \Omega$ . The students may arrange the cells in $m$ equal rows, each row holding $n$ cells in series, and then join the $m$ rows in parallel across the coil. Every cell must be used, so $m \times n = 12$ .

The group first tries the two simple arrangements: one row of all the cells, and the cells all in parallel. The teacher then asks them to find the arrangement that sends the largest current through the coil, and to give its current and power. Take the connecting wires as having no resistance. Remember that cells in series add their emfs and their internal resistances, and that $m$ identical rows in parallel divide the internal resistance of one row by $m$ .

What current flows through the coil when all the cells are in one row?
What current flows through the coil when all the cells are in parallel?
How many cells are there in each row of the best arrangement?
What is the largest current through the coil?
What power does the coil then receive?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Method. For $m$ rows of $n$ cells, the battery has emf $nE$ and internal resistance $\dfrac{nr}{m}$ . So $I = \dfrac{nE}{R + nr/m}$ , with $m = \dfrac{12}{n}$ .

(a) One row, $n = 12$ , $m = 1$ . $I = \dfrac{12 \times 1.5\ \text{V}}{2.0\ \Omega + 12 \times 0.50\ \Omega} = 2.25\ \text{A}$ .

(b) All in parallel, $n = 1$ , $m = 12$ . $I = \dfrac{1.5\ \text{V}}{2.0\ \Omega + 0.50\ \Omega / 12} = 0.73\ \text{A}$ .

(c) and (d) All the arrangements. Currents in amperes for $n = 1, 2, 3, 4, 6, 12$ : 1: 0.73, 2: 1.38, 3: 1.89, 4: 2.25, 6: 2.57, 12: 2.25. The largest is for $n = 6$ , with $m = 2$ rows, and the current is $2.57\ \text{A}$ . Here the battery has emf $9.0\ \text{V}$ and internal resistance $1.50\ \Omega$ , which is the nearest the arrangements come to the coil resistance of $2.0\ \Omega$ .

(e) Power. $P = I^2 R = (2.57\ \text{A})^2 \times 2.0\ \Omega = 13.2\ \text{W}$ .

Answer: (a) $2.25\ \text{A}$ . (b) $0.73\ \text{A}$ . (c) $6$ . (d) $2.57\ \text{A}$ . (e) $13.2\ \text{W}$ .

Check. The current is largest when the internal resistance of the battery matches the coil. Setting $\dfrac{nr}{m} = R$ with $m = \dfrac{12}{n}$ gives $n = \sqrt{\dfrac{12 \times 2.0}{0.50}} = 6.9$ . The whole numbers that divide $12$ nearest to this are 6 and 12, and 6 gives the larger current.

Watch out. More cells in series is not always better. With all $12$ in one row the internal resistance grows to $6.0\ \Omega$ and wastes most of the emf. The two simple arrangements are both worse than the mixed one, and the all-parallel one has the same emf as a single cell and delivers only $1.08\ \text{W}$ .

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